Playfield Arcade

FIXED PRACTICE BOARD

Bridges study 8

Solve at your own pace. Ask what follows from the position before applying the suggested move.

Study ID: bridges-08 · Original board: bridges-2027-05-13

Bridge networkA–B: 0 or 1 or 2 bridges; C–B: 0 or 1 or 2 bridges; D–A: 0 or 1 or 2 bridges; E–C: 0 or 1 or 2 bridges; F–D: 0 or 1 or 2 bridges; F–G: 0 or 1 or 2 bridges; H–F: 0 or 1 or 2 bridgesA3B3C3D4E1F4G1H1

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Work from the current board

Use the route buttons beneath the diagram. They cycle unknown (0, 1 or 2) → exactly 1 → exactly 2 → 0 → at least 1 (1 or 2) → unknown. Dotted lines are possible routes, not established bridges.

Tab to any control and use Enter or Space.

Interactive help is separate from the written starting-board solution below.

Review the techniques

Rules and proof limits

Join nearest visible islands horizontally or vertically with zero, one or two bridges per route. A bridge cannot pass through an island or cross another bridge between islands. Each island’s number counts individual bridges, and all islands must belong to one connected network.

A visible neighbour is only a possible route. Its remaining capacity and crossings may make it unusable. Correct island totals are insufficient without connectedness. Local capacity deductions need not finish every board; do not call a connection forced simply because it creates a tidy network.

Show the checked route from the starting position (12 steps)

This route starts from the original study position. It does not describe moves you made differently. Use “Explain a next step” for a deduction from your current marks.

  1. Removing A–B would split the graph of all still-possible routes. Every island must belong to one network, so A–B must carry at least one bridge. This does not by itself decide between one and two.
  2. Removing C–B would split the graph of all still-possible routes. Every island must belong to one network, so C–B must carry at least one bridge. This does not by itself decide between one and two.
  3. Removing D–A would split the graph of all still-possible routes. Every island must belong to one network, so D–A must carry at least one bridge. This does not by itself decide between one and two.
  4. Removing E–C would split the graph of all still-possible routes. Every island must belong to one network, so E–C must carry at least one bridge. This does not by itself decide between one and two.
  5. Removing F–D would split the graph of all still-possible routes. Every island must belong to one network, so F–D must carry at least one bridge. This does not by itself decide between one and two.
  6. Removing F–G would split the graph of all still-possible routes. Every island must belong to one network, so F–G must carry at least one bridge. This does not by itself decide between one and two.
  7. Removing H–F would split the graph of all still-possible routes. Every island must belong to one network, so H–F must carry at least one bridge. This does not by itself decide between one and two.
  8. Island D needs 4 individual bridges. Its routes currently allow 2 to 4 in total. Holding the other routes within their limits restricts D–A to 2; F–D to 2. A domain “1 or 2” proves a connection but does not yet choose single or double.
  9. Island A needs 3 individual bridges. Its routes currently allow 3 to 4 in total. Holding the other routes within their limits restricts A–B to 1. A domain “1 or 2” proves a connection but does not yet choose single or double.
  10. Island B needs 3 individual bridges. Its routes currently allow 2 to 3 in total. Holding the other routes within their limits restricts C–B to 2. A domain “1 or 2” proves a connection but does not yet choose single or double.
  11. Island C needs 3 individual bridges. Its routes currently allow 3 to 4 in total. Holding the other routes within their limits restricts E–C to 1. A domain “1 or 2” proves a connection but does not yet choose single or double.
  12. Island F needs 4 individual bridges. Its routes currently allow 4 to 6 in total. Holding the other routes within their limits restricts F–G to 1; H–F to 1. A domain “1 or 2” proves a connection but does not yet choose single or double.
Bridge networkA–B: 1 bridge; C–B: 2 bridges; D–A: 2 bridges; E–C: 1 bridge; F–D: 2 bridges; F–G: 1 bridge; H–F: 1 bridgeA3B3C3D4E1F4G1H1

Final check

Every island has its exact bridge count, routes do not cross, and a graph traversal reaches every island.

Back to the centreTry the regular game

Playfield Arcade · Updated 3 October 2026 · AI-assisted writing and code; examples checked against explicit constraints. Our method · Report a correction